Question

Estimate the value of the equilibrium constant at 690 K for each of the following reactions. ΔH∘fand S∘ for BrCl(g) is 14.6 kJ/mol and 240.0 Jmol⋅K, respectively.

Part A

2NO2(g)⇌N2O4(g)

Part B

Br2(g)+Cl2(g)⇌2BrCl(g)

Answer 1

Temperature = 690 K

Part A

2NO2(g)⇌N2O4(g)

ΔH∘f (NO2) = 33.8 kJ/mol

ΔH∘f (N2O4) = 9.2 kJ/mol

So (NO2) = - 60.8 J/mol-K

So (N2O4) = - 299 J/mol-K

Hof (rxn) = 9.2 - 2*33.8

= - 58.4 kJ/mol

So (rxn) = - 299 - (2*- 60.8)

= - 177.4 J/mol-K

ln Keq = 58.4 x 10³ / 8.314*690 - 177.4 / 8.314

= - 10.18 - 21.33

ln Keq = - 31.51

Keq = 2.067 x 10^-14

Part B

Br2(g)+Cl2(g)⇌2BrCl(g)

ΔH∘f (BrCl) = 14.6 kJ/mol

So (BrCl) = 240.0 Jmol⋅K

Hof (Br2) = 31 kJ/mol

Hof (Cl2) = 121.3 kJ/mol

So (Br2) = 245.46 J/mol-K

So (Cl2) = 223.08 J/mol-K

Hof (rxn) = 2*14.6 - (31 + 121.3)

= - 123.1 kJ/mol

So (rxn) = 2*240 - (245.46 + 223.08)

= 11.46 J/mol-K

ln Keq = - 123.1 x 10³ / 8.314*690 + 11.6 /8.314

= - 21.458 + 1.395

ln Keq = - 20.063

Keq = 1.93 x 10^-9